$sinx-\sqrt[]{3}cosx=2sin\dfrac{\pi}{4}$
$⇔ sinx-\sqrt[]{3}cosx=2sin\dfrac{\pi}{4}$
$⇔ \dfrac{1}{2}sinx-\dfrac{\sqrt[]{3}}{2}cosx=sin\dfrac{\pi}{4}$
$⇔ sinx.cos\dfrac{\pi}{3}-cosx.sin\dfrac{\pi}{3}=sin\dfrac{\pi}{4}$
$⇔ sin\Bigg(x-\dfrac{\pi}{3}\Bigg)=\dfrac{\pi}{4}$
$⇔ \left[ \begin{array}{l}x-\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=\dfrac{7\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{array} \right.$ $(k∈Z)$
